Answer:
Answer in explanation
Explanation:
The reason for the big size and less moisture of the cake is due to difference in weight of the ingredients on the surface of moon. So, the same has the lesser weight on the surface of the moon than it has on the surface of earth. Or in other words, The same weight of the ingredients will have greater mass and thus the greater quantity on the surface of earth than the surface of earth. For example, on earth 1.25 N butter will have a mass:
m = W/g = 1.25 N/(9.8 m/s²) = 0.13 kg
But, on moon:
m = W/g = 1.25 N/(1.625 m/s²) = 0.77 kg
Hence, it is clear that the mass of the same weight of the substance becomes 6 times greater on the surface of moon. This explains why the cake was so big.
Now, coming to the second part about the dryness of the cake. The main and only source of moisture in recipe is the eggs bu the eggs are taken in a quantity of numbers. So they are exactly the same on moon as well. While all the other ingredients are increased, the same amount of eggs are not sufficient to provide them with enough moisture. Hence the cake was dry.
The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct
Answer:
B) the particle's momentum.
Explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
[tex]F_c=\dfrac{m\times V^2}{R}[/tex]
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
[tex]F_m=q\times V\times B[/tex]
At the equilibrium condition
[tex]F_m=F_c[/tex]
[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]
[tex]R=\dfrac{m\times V}{q\times B}[/tex]
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.
Your professor is conducting a chemical isotope analysis of diet for her latest paleoanthropology project. Which of the following is true regarding isotopic analysis of diet?
a. Carbon analysis of bones and teeth will give insight to the types of plant material the hominin consumed.
b. Strontium analysis of bones and teeth will provide information regarding whether the hominin was a meat-eater.
c. Carbon analysis of bones and teeth will enable your professor to determine whether the meat consumed was from land or water sources.
d. Nitrogen analysis of bones and teeth will allow your professor to determine whether the hominin relied on C3 or C4 plants.
Answer:
Carbon analysis of bones and teeth will give insight to the types of plant material the hominin consumed.
Explanation:
This is because carbon being the principal composition of plants, analysis of the teeth's and bones for carbon will give an insight of what the homini n consumed
A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is located 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 kg. What is the tension in the back muscle
Answer:
T = 2689.6N
Explanation:
Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree
while both weight are acting perpendicular to the length. Hence we have :
Torque ( clockwise) = Torque ( anticlockwise)
m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1
Where m1 = 36kg
m2 = 20kg
g = 9.81m/s^2
Theta = 12
Substituting into equation 1
36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)
353.16L/2+196.2L = T ×0.2079(2L/3)
176.58L+196.2L = T × 0.1386L
372.78L = 0.1386LT
T = 372.78L/0.1386L
T = 2689.6N
Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen 77.0 cm behind the grating. First-order maxima are observed at distances of 58.0 cm , 65.4 cm , and 94.5 cm from the central maximum. What are the wavelengths of light emitted by element X?
Answer:
500 nm, 530 nm, 650 nm
Explanation:
Let's say that there is diffraction grating observed with a slit spacing of s. Respectively we must determine the angle θ which will help us determine the 3 wavelengths ( λ ) of the light emitted by element X. This can be done applying the following formulas,
s( sin θ ) = m [tex]*[/tex] λ, such that y = L( tan θ ) - where y = positioning, or the distance of the first - order maxima, and L = constant, of 77 cm
Now the grating has a slit spacing of -
s = 1 / N = 1 / 1200 = 0.833 [tex]*[/tex] 10⁻³ mm
The diffraction angles of the " positionings " should thus be -
θ = tan⁻¹ [tex]*[/tex] ( 0.58 / 0.77 ) = 37°,
θ = tan⁻¹ [tex]*[/tex] ( 0.654 / 0.77 ) = 40°,
θ = tan⁻¹ [tex]*[/tex] ( 0.945 / 0.77 ) = 51°
The wavelengths of these three bright fringes should thus be calculated through the formula : λ = s( sin θ ) -
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 37° ) = ( 500 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 40° ) = ( 530 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 51° ) = ( 650 [tex]*[/tex] 10⁻⁹ m )
Wavelengths : 500 nm, 530 nm, 650 nm
This question will be solved using the "grating equation".
The wavelengths of the light emitted by element X are:
"1. 6.654 x 10⁻⁷ m = 665.4 nm
2. 6.349 x 10⁻⁷ m = 634.9 nm
3. 5.262 x 10⁻⁷ m = 526.2 nm"
The diffraction grating equation is given as follows:
[tex]m\lambda = d Sin\ \theta[/tex]
where,
m = order of maxima = 1
λ = wavelength of light = ?
d = grating element = [tex]\frac{1}{no.\ of\ slits\ per\ unit\ length} = \frac{1}{1200\ slits/mm}[/tex]
d = (8.33 x 10⁻⁴ mm/slit)(1 m/ 1000 mm) = 8.33 x 10⁻⁷ m/slit
θ = angle of diffraction = [tex]tan^{-1}(\frac{L}{y})[/tex]
where,
L = distance of grating from the screen = 77 cm
y = distance of maxima from central maxima
Hence, the general equation after substituting constant values becomes:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{y}))[/tex]
FOR y = 58 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{58\ cm}))[/tex]
λ = 6.654 x 10⁻⁷ m = 665.4 nm
FOR y = 65.4 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{65.4\ cm}))[/tex]
λ = 6.349 x 10⁻⁷ m = 634.9 nm
FOR y = 94.5 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{94.5\ cm}))[/tex]
λ = 5.262 x 10⁻⁷ m = 526.2 nm
The attached picture shows the arrangement of the light rays in a diffraction grating.
Learn more about diffraction grating here:
https://brainly.com/question/17012571?referrer=searchResults
21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
I got it right on my quiz!
I hope this helps!! :))
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary
Answer:
1.32 is the turns ratio
Explanation:
Note that The transformer steps up the voltage from 12000 V to 390000V
12000 V is the primary and in the secondary it is 390000 V in old transformer
If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils
the formula is
n₂ / n₁ = voltage in secondry / voltage in primary
n₂ / n₁ = 390000 / 12000
ratio of turns in old transformer is 32.5
ratio of turns in new transformer
n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )
= 42.9
T he ratio of turns in the new secondary compared with the old secondary
n₃ / n₂ = 42.9 / 32.5
= 1.32
Light from an argon laser strikes a diffraction grating that has 4,917 lines per cm. The first-order principal maxima are separated by 0.4 m on a wall 1.62 m from the grating. What is the wavelength of the laser light in nm
Answer:
Wavelength is 4.8x10^-7m
Explanation:
See attached file
A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10
Answer:
It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water
Explanation:
. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio receiver. Explain how this is possible. Does it imply that TV audio is broadcast as FM
Answer:
Please see below as the answer is self-explanatory.
Explanation:
The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.
In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.
The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.
For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.
The reason why it is possible for TV station to sometimes pick up some of the audio portion on your FM radio receiver is because; TV waves can sometimes deviate into the FM radio frequency range.
Let us start with explaining the waves of TV and radio.
The frequency range utilized by TV stations is either the range 54 MHz to 88 MHz or 174 MHz to 222 MHz. In contrast, the frequency range utilized by FM Radio band is between 88 MHz and 174 MHz.
Now, in some cases, it is possible that the TV signal may deviate into the range of the FM Radio and as such in that case, the TV signal will pick the audio portion of an FM Radio. These TV waves are very high frequency waves.
Finally, it does not imply that the TV wave is broadcasting as an FM because it only deviated a bit from the TV range and not like that is where it is made to operate.
Read more about TV waves at; https://brainly.com/question/9684913
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
Ohm’s Law
pls answer this photos
Answer:
Trial 1: 2 Volts, 0 %
Trial 2: 2.8 Volts, 0%
Trial 3: 4 Volts, 0 %
Explanation:
Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:
V = IR
TRIAL 1:
V = IR
V = (0.1 A)(20 Ω)
V = 2 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2 - 2)/2| x 100%
% Difference = 0 %
TRIAL 2:
V = IR
V = (0.14 A)(20 Ω)
V = 2.8 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2.8 - 2.8)/2.8| x 100%
% Difference = 0 %
TRIAL 3:
V = IR
V = (0.2 A)(20 Ω)
V = 4 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(4 - 4)/4| x 100%
% Difference = 0 %
A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic friction between the rod and rails is 0.200, what vertical magnetic field is required to keep the rod moving at a constant speed?
Answer:
The magnetic field is [tex]B = 0.0764 \ T[/tex]
Explanation:
From the question we are told that
The mass of the metal is [tex]m = 0.210 \ kg[/tex]
The current is [tex]I = 11.0 \ A[/tex]
The distance between the rail(length of the rod ) is [tex]d = 0.490 \ m[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.200[/tex]
Generally the magnetic force is mathematically represented as
[tex]F_b = B * I * d[/tex]
Given that the rod is moving at a constant velocity, it
=> [tex]F_b = F_k[/tex]
Where [tex]F_k[/tex] is the kinetic frictional force which is mathematically represented as
[tex]F_k = \mu_k * m * g[/tex]
So
[tex]B * I * d = \mu_k * m * g[/tex]
=> [tex]B = \frac{\mu_k * m * g}{I * d }[/tex]
substituting values
=> [tex]B = \frac{0.200 * 0.210 * 9.8 }{ 11 * 0.490 }[/tex]
=> [tex]B = 0.0764 \ T[/tex]
1. Notice that the voltmeter moves in response to the coil entering or leaving the magnetic gap.
2. Let's apply Faraday's Law to this situation. Faraday's Law says that the induced voltage (or emf )in a loop of wire caused by a changing magnetic field is:
€ = 1
Where is the magnetic flux which is
Q = BA
In this case, the flux density B is not changing. Instead, the changing flux is due to the motion of the coil as it enters or leaves the magnetic gap:
do = BdA
Given that the area immersed in the gap is changing as the coil enters the gap, what is the correct expression of Faraday's Law for this situation?
Answer:
Explanation:
let the coil of length l and breathe b entering the magnetic field B with speed v.
So, the magnetic flux through the coil is
Ф = B(l×b)
length × breathe = area
Ф = BA
dФ = BdA
therefore induced emf is given as
ε = [tex]Bl(\frac{db}{dt})[/tex]
note: [tex]\frac{db}{dt} = v[/tex]
ε[tex]= Blv[/tex]
attached is the diagram for the solution
Recent technological developments like high-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have refined and extended the camera’s capacity to provide information. Which passage assertion does this information support most strongly?
Answer:
D) Photography can be used to both control and benefit society.
Explanation:
High-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have been used to both control and benefits the society in the sense that it has helped to take records of information of crime, traffic offenders such drunk drivers and over speeding drivers, e.t.c. it helps control by given their information and automatically penalizing them or ensuring the agency penalized them and also benefit the society by preventing people from committing crime thereby, protecting them from offenders.
Asteroid A has 3.5 times the mass and 2.0 times the velocity of Asteroid B. If
Asteroid B has a kinetic energy of 2,300,000 J then what is the kinetic energy of
Asteroid A?
Answer:
K_A = 32.2 10⁶ J
Explanation:
In this exercise we must relate the quantities given to find the kinetic energy
Asteroid A data
m_A = 3.5 m_B
v_A = 2.0 v
they also give the value of the kinetic energy of asteroid A
K_B = 2.3 10⁶ J
the expression for scientific energy is
K = ½ m v²
let's replace
K_A = ½ m_a V_a2
K_A = ½ 3.5 m_B (2.0 v_B)^2
K_A = 3.5 2² (½ m_B v_B²)
K_A = 14 K_B
K_A = 32.2 10⁶ J
two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of 2.01 kg of this oil from 23 °C to 191 °C?
Answer:
Q = 590,940 J
Explanation:
Given:
Specific heat (c) = 1.75 J/(g⋅°C)
Mass(m) = 2.01 kg = 2,010
Change in temperature (ΔT) = 191 - 23 = 168°C
Find:
Heat required (Q)
Computation:
Q = mcΔT
Q = (2,010)(1.75)(168)
Q = 590,940 J
Q = 590.94 kJ
Need help understanding this. If anyone help, that would be greatly appreciated!
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
a) acceleration:
ā=v^2/r
ā=(15m/s)^2/27m
ā=225/27 m/s^2
ā=8.333 m/s^2
force:
F=mā. where the is equal to v^2/r
F=1000kg*8.3 m/s^2
F=8333.3 N
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
λ = 5.2 x 10⁻⁷ m = 520 nm
I need help with this question?
Answer:
You got it right, didn't you?
c) vector C
Explanation:
opposite to vector V
Answer:
A, vector B
Explanation:
A negative vector is a vector which points in the opposite direction, even though it’s in a different quadranot it’s still the opposite direction.
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long.
(a) If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock?
(b) What is the mechanical advantage of this lever system?
Answer:
a
[tex]F =326.7 \ N[/tex]
b
[tex]M = 6[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 200 \ kg[/tex]
The length of the small object from the rock is [tex]d = 2 \ m[/tex]
The length of the small object from the branch [tex]l = 12 \ m[/tex]
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
[tex]W = m_r * g[/tex]
substituting values
[tex]W = 200 * 9.8[/tex]
[tex]W = 1960 \ N[/tex]
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
[tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]
Here [tex]\theta[/tex] is very small so [tex]cos\theta * l = l[/tex]
and [tex]cos\theta * d = d[/tex]
Hence
[tex]F * l - W * d = 0[/tex]
=> [tex]F = \frac{W * d}{l}[/tex]
substituting values
[tex]F = \frac{1960 * 2}{12}[/tex]
[tex]F =326.7 \ N[/tex]
The mechanical advantage is mathematically evaluated as
[tex]M = \frac{W}{F}[/tex]
substituting values
[tex]M = \frac{1960}{326.7}[/tex]
[tex]M = 6[/tex]
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done by gravity during the roll, in Joules
Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
[tex]F = W_{x} = mgsin(\theta)[/tex]
Where:
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]
Now, we can find the work:
[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point B.
Answer:
[similar to]
Explanation:
it is the missing word
An optical fiber uses one glass clad with another glass. What is the critical angle? (Assume the glass in the fiber has an index of refraction of 1.69, and the glass in the cladding has an index of refraction of 1.50.)
Answer:
The critical angle is [tex]\theta _c = 62.57^o[/tex]
Explanation:
From the question we are told that
The index of refraction of the glass in the fiber is [tex]n_f = 1.69[/tex]
The index of refraction of the glass in the cladding is [tex]n_c = 1.50[/tex]
The critical angle is mathematically evaluated as
[tex]\theta_c = sin^{-1}[\frac{n_c }{n_f } ][/tex]
substituting values
[tex]\theta_c = sin^{-1}[\frac{1.50 }{1.69 } ][/tex]
[tex]\theta _c = 62.57^o[/tex]
The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The spring scale [tex]F_2[/tex] reads [tex]F_2 = 2.4225 \ N[/tex]
Explanation:
From the question we are told that
The first force is [tex]F_1 = 10.5 \ N[/tex]
The acceleration by which the cart moves to the right is [tex]a = 2.50 \ m/s^2[/tex]
The mass of the cart is m = 3.231 kg
Generally the net force on the cart is
[tex]F_{net} = F_1 - F_2[/tex]
This net force is mathematically represented as
[tex]F_{net} = m * a[/tex]
So
[tex]m* a = 10 - F_2[/tex]
[tex]F_2 = 10.5 - 2.5 (3.231)[/tex]
[tex]F_2 = 2.4225 \ N[/tex]
When a current of 2.0 A flows in the 100-turn primary of an ideal transformer, this causes 14 A to flow in the secondary. How many turns are in the secondary
Answer:
14.29 turns.
Explanation:
From the question,
Ns/Np = Ip/Is........................ Equation 1
Where Ns = Secondary turn, Np = Primary turn, Is = current flowing in the secondary turn, Ip = current flowing in the primary turn.
Make Ns the subject of the equation
Ns = NpIp/Is.................... Equation 2
Given: Np = 100 turns, Ip = 2.0 A, Is = 14 A.
Substitute these values into equation 2
Ns = 100(2.0)/14
Ns = 14.29 turns.
A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe
Answer:
0.0127m
Explanation:
Using
Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m
Solve 3* +5-220t = 0
Answer:
t = 27.5
Explanation:
[tex]3 + 5 -220t = 0[/tex]
Well to solve for t we need to combine like terms and seperate t.
So 3+5= 8
8 - 220t = 0
We do +220 to both sides
8 = 220t
And now we divide 220 by 8 which is 27.5
Hence, t = 27.5
An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)
Answer:
The induced emf is [tex]\epsilon =7.68 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1300 \ turns[/tex]
The diameter is [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]
The initial magnetic field is [tex]B_i = 0.14 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
The time taken is [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2 }[/tex]
substituting values
[tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]
[tex]r = 1.1*10^{-2} \ m[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]
Where [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as
[tex]d\phi = dB * A * cos\theta[/tex]
=> [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]
Here [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field
Also A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]
[tex]A = 3.8 *10^{-4] \ m^2[/tex]
So
[tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]
[tex]d\phi = -5.32*10^{-5} \ weber[/tex]
So
[tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]
[tex]\epsilon =7.68 \ V[/tex]